We note first that division by zero is not allowed and that for this reason \(5x+9\) may not be equal to zero and that therefore \(x=-{{9}\over{5}}\) is not a solution.

We now distinguish two cases, namely \(5x+9>0\) and \(5x+9<0\).
In both cases we multiply the inequality on both sides by \(5x+9\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(5x+9>0\), i.e. \(x> -{{9}\over{5}}\). Then we get \(7<-6(5x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(30x<-61\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{61}\over{30}}\).
So we have the following system of inequalities: \(x> -{{9}\over{5}}\,\wedge\; x < -{{61}\over{30}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(5x+9<0\), i.e. \(x< -{{9}\over{5}}\). Then we get \(7>-6(5x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(30x>-61\).
Then, division by the coefficient of \(x\) gives \(x > -{{61}\over{30}}\).
So we have the following system of inequalities: \(x< -{{9}\over{5}}\,\wedge\; x > -{{61}\over{30}}\)
and this simplifies to \(-{{61}\over{30}}\lt x\lt -{{9}\over{5}}\).

The solution of the original inequality is \(-{{61}\over{30}}\lt x\lt -{{9}\over{5}}\).